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I Differentiation of sin function where's my mistake?

  1. Dec 21, 2017 #1
    I was thinking and came up with this. I know it's wrong but can't find the mistake :(

    dy/dx sin(x) = cos(x)
    dy/dx sin(kx) = kcos(kx)

    So dy/dx sin(3x) = 3cos(3x)
    Now let Y = 3x
    dy/dx sin(Y) = cos(Y) = cos(3x)

    3cos(3x) = cos(3x)
    3 = 1

    Where is the mistake?
  2. jcsd
  3. Dec 21, 2017 #2


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    ## \dfrac{d}{dx}\sin(Y) \neq \cos(Y)##
    You've made the chain rule vanish by pretending ##Y=y##.
  4. Dec 21, 2017 #3


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    Your notation doesn’t make sense.
    dy/dx is the derivative of y with respect to x. “dy/dx sin(x)” is not a well-defined expression.

    What you nean is d/dx sin(x). And suddenly the issue disappears:

    d/dx sin(3x)=3cos(3x)
    d/dx sin(y)=? - here we need the chain rule and d/dx 3x = 3:
    d/dx sin(y) = cos(y) d/dx y = cos(y) * 3 = 3cos(3x)
  5. Dec 21, 2017 #4
    I haven't done the chain rule yet I'll need to look into that :)
  6. Dec 21, 2017 #5


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    If y =3x changes three times as fast as x, and sin(y) changes cos(y) times as fast as y, then sin(y)=sin(3x) changes 3*cos(y) = 3*cos(3x) times as fast as x.
  7. Dec 21, 2017 #6


    Staff: Mentor

    As already noted, the above should be ##\frac d {dx}\left(\sin(3x)\right) = 3\cos(3x)##
    ##\frac{dy}{dx}## is the derivative of y with respect to x, so it is a thing, a noun.
    ##\frac d{dx}## is the operator that signifies taking the derivative of something with respect to x. It is an action that hasn't completed yet, a verb. Don't confuse these two things.

    Actually, you used the chain rule in the first line of what I quoted, above. The chain rule is what gives you that leading factor of 3.
  8. Dec 21, 2017 #7
    I don't know, I just used a table of derivatives :/
  9. Dec 21, 2017 #8


    Staff: Mentor

    They were using the chain rule in the table.
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